Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $t = \dfrac{x - 10}{-3x - 18} \times \dfrac{5x^3 - 95x^2 + 450x}{x^3 - 19x^2 + 90x} $
Answer: First factor out any common factors. $t = \dfrac{x - 10}{-3(x + 6)} \times \dfrac{5x(x^2 - 19x + 90)}{x(x^2 - 19x + 90)} $ Then factor the quadratic expressions. $t = \dfrac {x - 10} {-3(x + 6)} \times \dfrac {5x(x - 9)(x - 10)} {x(x - 9)(x - 10)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {(x - 10) \times 5x(x - 9)(x - 10) } {-3(x + 6) \times x(x - 9)(x - 10) } $ $t = \dfrac {5x(x - 9)(x - 10)(x - 10)} {-3x(x - 9)(x - 10)(x + 6)} $ Notice that $(x - 9)$ and $(x - 10)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {5x\cancel{(x - 9)}(x - 10)(x - 10)} {-3x\cancel{(x - 9)}(x - 10)(x + 6)} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $t = \dfrac {5x\cancel{(x - 9)}(x - 10)\cancel{(x - 10)}} {-3x\cancel{(x - 9)}\cancel{(x - 10)}(x + 6)} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $t = \dfrac {5x(x - 10)} {-3x(x + 6)} $ $ t = \dfrac{-5(x - 10)}{3(x + 6)}; x \neq 9; x \neq 10 $